## anonymous one year ago FAN+MEDAL+TESTIMONIAL:):)!! How many arrangements are there of the word MATHEMATICS? How many of these start with the letter M? How many of the arrangements in part a have the T’s together?

1. anonymous

I think that this is a permutation problem. So given that there are 11 letters in the word math, I will do the following calculation

2. anonymous

11P11

3. anonymous

Without restrictions

4. anonymous

@nincompoop

5. anonymous

hai

6. anonymous

Hi Sir:)

7. anonymous

Am I on the right track?

8. anonymous

ya know I'm stuck on this question right now too :P

9. anonymous

This is a university level math don't worryXD

10. anonymous

:O really? its on my 8th grade quiz

11. anonymous

Forgive me I am in a Canadian school.

12. anonymous

lol ok i will try to help tho

13. anonymous

gimee a sec

14. anonymous

Arrangement of letters of a word is a permutation question right?

15. anonymous

As opposed to combination

16. anonymous

Hi everyone:)

17. anonymous

Am I on the right track?

18. anonymous

I don't know.

19. anonymous

How many arrangements are there of the word MATHEMATICS? Rule: Start with the factorial of the number of letters in the word. Then, for each indistinguishable letter in the word, divide by the factorial of the number of times that letter occurs in the word. "MATHEMATICS" is an 11-letter word. If all the letters were distinguishable like in "MATHEmatICS", the answer would be 11! = 39916800 However, there are 2 indistinguishable M's 2 indistinguishable A's 2 indistinguishable T's Thus, using the rule, we divide 11! by (2!)(2!)(2!) $\frac{ 11! }{ 2!*2!*2! }=\frac{ 39916800 }{ 8 }=4989600$ How many of these start with the letter M? That amounts to finding all the distinguishable arrangements of the 10-letter "word" "ATHEMATICS" and putting an M in the beginning of each. "ATHEMATICS" is a 10-letter "word" and it contains 2 indistinguishable A's 2 indistinguishable T's Thus, using the rule, we divide 10! by (2!)(2!) $\frac{ 10! }{ 2!*2! }=\frac{ 3628800 }{ 4 }=907200$ How many of the arrangements in part a have the T’s together? That amounts to finding all the distinguishable arrangements of the 10-letter "word" "MATHEMAICS" and inserting another T to the right of the "T" in each. "MATHEMAICS" is a 10-letter "word" and it contains 2 indistinguishable M's 2 indistinguishable A's Thus, using the rule, it's exactly the same answer as the second part. We divide 10! by (2!)(2!) $\frac{ 10! }{ 2!*2! }=\frac{ 3628800 }{ 4 }=907200$ There ya go!

20. anonymous

YOU ROCK:)

21. anonymous

:)

22. anonymous

ok testimonial?

23. anonymous

let me know if it could be better